Question

Two lines $l_{1}x+m_{1}y+n_1=0$ and $l_{2}x+m_{2}y+n_2=0$ are conjugate lines with respect to the circle $x^2+y^2=a^2$, if

• A. $l_1{l}_2+m_1{m}_2+n_1{n}_2=0$
• B. $l_1{l}_2+m_1{m}_2=n_1{n}_2$
• C. $a^{2(}{l}_1{l}_2+m_1{m}_2)=n_1{n}_2$
• D. $l_1{l}_2+m_1{m}_2=a^2{n}_1{n}_2$

Answer 1

The correct option is C $a^{2(}{l}_1{l}_2+m_1{m}_2)=n_1{n}_2$

Two lines are said to be conjugate w.r.t to a circle if one line passes through pole of the other line.

i.e, $l_{2}x+m_{2}y+n_2$ passes through pole of $l_{1}x+m_{1}y+n_1=0$

Given, circle eq'n is $x^2+y^2=a^2$

Center of the circle is $(0,0)$

Foot of the perpendicular from origin to the line $l_{1}x+m_{1}y+n_1=0$ is $A^{\prime }(\frac{-n_1{l}_1}{l_1^2+m_1^2},\frac{-n_1{m}_1}{l_1^2+m_1^2})$

Distance of A' from center is OA'$=\sqrt{\frac{n_1^2}{l_1^2+m_1^2}}$

Let the pole of this line be A.

Then , $OA.O{A}^{\prime }=a^2$

$O{A}^2.O{A}^{\prime 2}=a^4$

$O{A}^2=\frac{a^4(l_1^2+m_1^2)}{n_1^2}$

Let $A=(x,y)$

Then, $x^2+y^2=\frac{a^4(l_1^2+m_1^2)}{n_1^2}...\left(1\right)$

$A,O,A^{\prime }$ are collinear.

Therefore, $\frac{x}{y}=\frac{l_1}{m_1}$

$y=\frac{m_1}{l_1}x$

substituting in (1) ,

$x^2=\frac{a^4{l}_1^2}{n_1^2},x=\pm \frac{a^2{l}_1}{n_1},y=\pm \frac{a^2{m}_1}{n_1}$

For the line line $l_{2}x+m_{2}y+n_2=0$ to be conjugate of $l_{1}x+n_{1}y+n_1$, it should pass through A.

i.e $l_2(\pm \frac{a^2{l}_1}{n_1})+m_2(\pm \frac{a^2{m}_1}{n_1})+n_2=0$

$=a^2(l_1{l}_2+m_1{m}_2)=\mp n_1{n}_2$